Complete and balance each gas-evolution equation. a. HNO3(aq) + Na2SO3(aq)¡

Video transcript

Welcome back everyone. The reaction between Aquarius hydrogen fluoride and Aquarius strontium sulfide. It produces a gas right? The balanced equation for this gas evolution reaction were given our reactant hydrogen fluoride and strontium sulfide. They're both Aquarius. And in order to predict our products, we're going to recognize that this is going to be a double replacement for displacement reaction. So recall that in a double replacement reaction, our cat ions will combine with one another. Or sorry, our captain will combine with an ion. So in the case of our H plus kati on, it's going to combine with our cell fight an ion. Recall that sulfide as a poly atomic ion where we'll have the two minus an ion charge and our hydrogen ion is a H plus catalon. And that's because hydrogen we recall is in group one A on the periodic table and any atom in group one A will form a plus one ion charge. Next we'll combine our fluoride. An ion with our strontium caddy on where we recall because florian is also a group seven A ion, it's going to form a minus one charge. So we would have f minus one and strong Tm. We recognize as a group to a on the periodic table and will form a two plus charge as a caddy on. So now combining our ions will form our products with the combination of H plus and S. +032 will have a church sub two since hydrogen will get the charge of our cell fight an ion and then we'll have S. +03 because sulfite will get the charge of our hydrogen ion caddy on this is going to be our first Aquarius product as sulfurous acid. And then for our second product, we're going to form the combination of our fluoride an ion and our strontium carry on, which is going to be S. R. Sub one, since strontium will get the anti on charge of flooring as one and then F sub two since flooring will get the caddy on charge on strontium. And this is also an Aquarius product because it's going to dissociate into it's ions. Now for our sulfurous acid product, we want to recognize that sulfurous acid is actually thermo dynamically and stable in this state. And so because it is well known to be a thermo dynamically unstable product here, we're going to have to recall that it's going to spontaneously decompose into water and sulfur dioxide gas where it would be more stable since these products have lower energy than sulfurous acid. So we're going to need to also make sure that our reaction is balanced with these products. And if we recognize that on the product side, we have two moles of hydrogen and two moles of flooring, we're going to need that on the reactant side by placing a coefficient of two in front of our hydrogen fluoride, which also will give us two moles of hydrogen and two moles of flooring. So now with this coefficient added and with knowing that we need to spontaneously decomposed are sulfurous acid into water and sulfur dioxide gas. That's also going to help us answer this prompt since we need a gaseous product. So rewriting our entire equation, what we should have is two moles of hydrogen fluoride Reacting with one mole of strontium sulfide, which will produce water plus sulfur dioxide gas, plus our strontium fluoride product that is Aquarius. And this entire equation with our coefficient is going to complete this example as our final answer for our bounced reaction and our gaseous product sulfurous acid. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.